Abstract Algebra Theory and Applications - Computer Science ...

20.2 POLYNOMIAL CODES 353f(x) = a 0 + a 1 x + a 2 x 2 by 1 + x 3 . We are defining a map φ : Z 3 2 → Z6 2by φ : f(x) ↦→ g(x)f(x). It is easy to check that this map is a grouphomomorphism. In fact, if we regard Z n 2 as a vector space over Z 2, φ is alinear transformation of vector spaces (see Exercise 13, Chapter 18). Letus compute the kernel of φ. Observe that φ(a 0 , a 1 , a 2 ) = (000000) exactlywhen0 + 0x + 0x 2 + 0x 3 + 0x 4 + 0x 5 = (1 + x 3 )(a 0 + a 1 x + a 2 x 2 )= a 0 + a 1 x + a 2 x 2 + a 0 x 3 + a 1 x 4 + a 2 x 5 .Since the polynomials over a field form an integral domain, a 0 + a 1 x + a 2 x 2must be the zero polynomial. Therefore, ker φ = {(000)} and φ is one-to-one.To calculate a generator matrix for C, we merely need to examine theway the polynomials 1, x, and x 2 are encoded:(1 + x 3 ) · 1 = 1 + x 3(1 + x 3 )x = x + x 4(1 + x 3 )x 3 = x 2 + x 5 .We obtain the code corresponding to the generator matrix G 1 in Example 4.The parity-check matrix for this code is⎛1 0 0 1 0⎞0H = ⎝ 0 1 0 0 1 0 ⎠ .0 0 1 0 0 1Since the smallest weight of any nonzero codeword is 2, this code has theability to detect all single errors.Rings of polynomials have a great deal of structure; therefore, our immediategoal is to establish a link between polynomial codes and ring theory.Recall that x n − 1 = (x − 1)(x n−1 + · · · + x + 1). The factor ringR n = Z 2 [x]/〈x n − 1〉can be considered to be the ring of polynomials of the formf(t) = a 0 + a 1 t + · · · + a n−1 t n−1that satisfy the condition t n = 1. It is an easy exercise to show that Z n 2 andR n are isomorphic as vector spaces. We will often identify elements in Z n 2with elements in Z[x]/〈x n − 1〉. In this manner we can interpret a linearcode as a subset of Z[x]/〈x n − 1〉.